# A cardinal inequality for finiteness

Nearly ten years ago, I explained in a blog post that, assuming only ZF, a cardinal number $$\mathfrak{n}$$ is finite if and only if it satisfies this monstrous inequality: $$2^{2^{2^{2^{\mathfrak{n}}}}} \lt \left(2^{2^{2^{2^{\mathfrak{n}}}}}\right)^2 = 4^{2^{2^{2^{\mathfrak{n}}}}}$$ Where "finite" is meant in the strictest Tarskian sense: in bijection with a finite ordinal.

My question is a bit vague since "simple" has a lot of interpretations but here it is:

Is there a simpler cardinal inequality that is equivalent to finiteness and uses only cardinal exponentiation?

• Oh that's hideous, I love it! Commented May 22 at 15:09
• I suspect 4 could be replaced by 3 but even that requires some serious work. I expect the answer to be negative in some sense but I don't know what that "sense" is. Commented May 22 at 15:32

Läuchli proved in 1961 that $$\mathfrak{n}$$ is finite if and only if $$2^{2^{\mathfrak{n}}}＜2^{2^{\mathfrak{n}}}\cdot2$$. As a consequence, $$\mathfrak{n}$$ is finite if and only if $$2^{2^{2^{\mathfrak{n}}}}＜(2^{2^{2^{\mathfrak{n}}}})^2$$. It is still open (asked by Läuchli) whether $$\mathfrak{n}$$ is finite if and only if $$2^{2^{\mathfrak{n}}}＜(2^{2^{\mathfrak{n}}})^2$$.